Vector subspace

A vector subspace is a subset of a vector space that is still closed under addition and scalar multiplication.

Definition

Let V be a vector space. A vector subspace W is then a subset of V that is itself a vector space. A non-empty subset W ⊆ V must therefore satisfy the following two conditions to be a vector subspace of V: for all x, y ∈ W and for every a ∈ ℝ:

• x + y ∈ W,
• a · x ∈ W.

So we have to choose a subset of V whose vectors are closed under addition and multiplication.

A subspace of R³

Let's take the space ℝ³ and look for a subspace.

1. What about all triples of the form (a, a, a), where a ∈ ℝ? That is, triples such as $(1,1,1),(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ or (−π, −π, −π). Such a subspace, which we call W₁, would be a subset of ℝ³, that is, W₁ ⊆ ℝ³, because ℝ³ contains all triples.

Such a space would be closed under vector addition, because

$$(a,a,a)+(b,b,b)=(a+b,a+b,a+b).$$

After adding two vectors we get a new vector with three equal entries again, and such a vector belongs to the set W₁. So the set W₁ satisfies the first condition for a vector subspace. But does it satisfy the second condition?

Yes, it does, because multiplying by k changes all three components of the vector, and changes them in exactly the same way:

$$k\cdot (a,a,a)=(k\cdot a,k\cdot a,k\cdot a).$$

Again we get a vector with all three components equal, and such a vector is in the set W₁. So the set W₁ is a vector subspace of ℝ³.

• We could try taking all triples (a, b, c) such that a, b, c ∈ [−1, 1]. We call this set W₂, and its elements include, for example, the triples $(\frac{1}{2},-\frac{2}{7},\frac{8}{9})$ and $(0,\frac{\pi }{4},1)$. Are the elements of W₂ closed under addition? Definitely not, because $(1,1,1)+(\frac{1}{2},\frac{1}{3},\frac{1}{4})$ is equal to the triple $(\frac{3}{2},\frac{4}{3},\frac{5}{4})$, which does not belong to W₂. So the set W₂ is not a subspace of ℝ³; in fact, the set W₂ is not a vector space at all.

• Now let's take all triples of the form (a, 0, b), where a, b ∈ ℝ. These are all the triples whose second component is zero. We call this set W₃. Is this set closed under addition? Yes, it is, because

$$(a,0,b)+(c,0,d)=(a+c,0,b+d)$$

The result is the triple (a + c, 0, b + d), whose second component is zero and whose other two components are real numbers. This triple is certainly in W₃. Is W₃ closed under multiplication? Yes, it is, because

$$k\cdot (a,0,b)=(k\cdot a,0,k\cdot b)$$

Again we get the triple (k · a, 0, k · b), whose second component is always zero and whose other components are real numbers. Such a triple is an element of W₃. So the set W₃ forms a subspace of ℝ³. Notice that if the triples had the form (a, b, 0) or (a, 0, 0), they would also form a vector subspace.

A subspace of polynomials

In the previous article we showed a vector space made up of polynomials. To recap: a polynomial is an expression of the form

$$p(x)=a_0+a_{1}x+a_2{x}^2+\dots +a_n{x}^n.$$

We assume that a_n ≠ 0. The degree of such a polynomial is then exactly the number n. The real numbers a₀, …, a_n are called the coefficients of the polynomial. An example of a polynomial is the expression 4 + 3x − 7x². Let's try to find a vector subspace. Let P(x) be the vector space of all polynomials.

1. First let's try the set of all polynomials of degree n, where n ≥ 0, and call this set W₁. In the previous article we already said that this is not a vector space at all, so it is hardly going to be a vector subspace of P(x). Let's repeat why: if we multiply a polynomial in W₁ by zero, we get the polynomial p(x) = 0, which by definition has degree −1. So this element is certainly not in any W₁.

2. For W₂ let's try taking all polynomials of degree n or lower. So for n = 2, W₂ contains all polynomials of degree 2, 1, 0, −1. Is this set closed under addition? Yes, it is, because the sum of two polynomials of degree n never has a degree greater than n; it can only be lower. This set is also closed under multiplication, because any scalar multiple k · p(x) of a polynomial p(x) is either a polynomial of the same degree or the zero polynomial. The set W₂ covers both possibilities, so W₂ forms a subspace of P(X).

Basic properties of subspaces

Let V be a vector space and W₁ and W₂ two of its subspaces. Then the following hold:

1. The intersection W₁ ∩ W₂ is a vector subspace of V.

   Take the vector space ℝ³ and subspaces W₁ and W₂ such that W₁ consists of triples of the form (a, b, 0) and W₂ of triples of the form (a, 0, b), where a, b ∈ ℝ. The intersection W₁ ∩ W₂ is then the set containing triples of the form (a, 0, 0) – such a set is clearly a subspace of ℝ³.

   Of course, that is not a proof. A proof would look like this: take any vector space V and two of its subspaces W₁ and W₂. We want to prove that for all x, y ∈ W₁ ∩ W₂ we have x + y ∈ W₁ ∩ W₂.

   Since the vectors x and y are in the intersection W₁ ∩ W₂, it must be true that x, y ∈ W₁ and also x, y ∈ W₂ (this follows from the definition of the intersection of sets). Since W₁ and W₂ are vector spaces, it must also hold that x + y ∈ W₁ and also x + y ∈ W₂ (this follows from the definition of vector spaces). But now we have x + y ∈ W₁ and also x + y ∈ W₂, which implies that x + y ∈ W₁ ∩ W₂ (again from the definition of the intersection of sets – if an element lies in the set W₁ and also in the set W₂, it must lie in their intersection).

   Next we must check closure under multiplication. We must prove that for all k ∈ ℝ and x ∈ W₁ ∩ W₂ we have k · x ∈ W₁ ∩ W₂. The procedure is the same. Since x ∈ W₁ ∩ W₂, we have both x ∈ W₁ and x ∈ W₂. Since W₁ and W₂ are spaces, k · x ∈ W₁ and also k · x ∈ W₂, and therefore k · x ∈ W₁ ∩ W₂. □

2. The union W₁ ∪ W₂ is not necessarily a subspace of V.

   To prove the second property, it is enough to find a suitable counterexample. We can make do with the previous spaces W₁ and W₂, which have triples of the form (a, b, 0) and (a, 0, b) respectively. If we take their union, we get all the triples that must have a zero in the second or third place. So the vectors (1, 0, 1) and (1, 1, 0) are certainly in the union W₁ ∪ W₂. But their sum is the vector (2, 1, 1), which has no zero anywhere. This vector is in neither W₁ nor W₂, so it cannot be in W₁ ∪ W₂. So we have found two vectors whose sum does not belong to the same set, which means that the union cannot be a vector space.